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Solve for $x$: 1.1.1 $-2x(x + a)(3 - x) = 0$ 1.1.2 $2x = 6 - x^2$ (correct to TWO decimal places) 1.1.3 $5x(x - 3) \, <= 0$ and then represent the solution on a number line - NSC Technical Mathematics - Question 1 - 2018 - Paper 1

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Question 1

Solve-for-$x$:--1.1.1---$-2x(x-+-a)(3---x)-=-0$----1.1.2---$2x-=-6---x^2$---(correct-to-TWO-decimal-places)--1.1.3---$5x(x---3)-\,-<=-0$-and-then-represent-the-solution-on-a-number-line-NSC Technical Mathematics-Question 1-2018-Paper 1.png

Solve for $x$: 1.1.1 $-2x(x + a)(3 - x) = 0$ 1.1.2 $2x = 6 - x^2$ (correct to TWO decimal places) 1.1.3 $5x(x - 3) \, <= 0$ and then represent the solut... show full transcript

Worked Solution & Example Answer:Solve for $x$: 1.1.1 $-2x(x + a)(3 - x) = 0$ 1.1.2 $2x = 6 - x^2$ (correct to TWO decimal places) 1.1.3 $5x(x - 3) \, <= 0$ and then represent the solution on a number line - NSC Technical Mathematics - Question 1 - 2018 - Paper 1

Step 1

1.1.1 Solve for $x$: $-2x(x + a)(3 - x) = 0$

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Answer

To solve the equation, we can set each factor to zero:

  1. 2x=0x=0-2x = 0 \Rightarrow x = 0
  2. (x+a)=0x=a(x + a) = 0 \Rightarrow x = -a
  3. (3x)=0x=3(3 - x) = 0 \Rightarrow x = 3

Thus, the solutions are x=0x = 0, x=ax = -a, and x=3x = 3.

Step 2

1.1.2 Solve for $x$: $2x = 6 - x^2$ (correct to TWO decimal places)

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Answer

Rearranging the equation gives us:

x2+2x6=0x^2 + 2x - 6 = 0

Using the quadratic formula, where a=1a = 1, b=2b = 2, and c=6c = -6:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the values, we find:

x=2±4+242=2±282=2±272=1±7x = \frac{-2 \pm \sqrt{4 + 24}}{2} = \frac{-2 \pm \sqrt{28}}{2} = \frac{-2 \pm 2\sqrt{7}}{2} = -1 \pm \sqrt{7}

Calculating the approximations:

  1. x1.65x \approx 1.65
  2. x3.65x \approx -3.65 (Not valid in this context.)

Therefore, the accepted value is x1.65x \approx 1.65.

Step 3

1.1.3 Solve for $x$: $5x(x - 3) \leq 0$

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Answer

To analyze the inequality 5x(x3)05x(x - 3) \leq 0, first find the critical values:

  1. x=0x = 0
  2. x=3x = 3

Next, we test intervals created by these critical values:

  1. For x<0x < 0: (choose x=1x = -1) → 5(1)(13)=20>05(-1)(-1 - 3) = 20 > 0
  2. For 0<x<30 < x < 3: (choose x=1x = 1) → 5(1)(13)=10<05(1)(1 - 3) = -10 < 0
  3. For x>3x > 3: (choose x=4x = 4) → 5(4)(43)=20>05(4)(4 - 3) = 20 > 0

Thus, the solution is x[0,3]x \in [0, 3], which should be represented on a number line.

Step 4

1.2 Solve for $x$ and $y$: $y - 2x = -7$ and $x^2 + xy + y^2 = 21$

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Answer

From y2x=7y - 2x = -7, rearranging gives: y=2x7y = 2x - 7

Substituting this into x2+xy+y2=21x^2 + xy + y^2 = 21:

x2+x(2x7)+(2x7)2=21x^2 + x(2x - 7) + (2x - 7)^2 = 21

Expanding and simplifying:

tx2+2x27x+4x228x+49=21x^2 + 2x^2 - 7x + 4x^2 - 28x + 49 = 21 7x235x+28=07x^2 - 35x + 28 = 0

Dividing through by 7: x25x+4=0x^2 - 5x + 4 = 0

Factoring gives: (x4)(x1)=0(x - 4)(x - 1) = 0

Thus, x=4x = 4 or x=1x = 1. Plugging these back into y=2x7y = 2x - 7:

  • For x=4x = 4: y=2(4)7=1y = 2(4) - 7 = 1
  • For x=1x = 1: y=2(1)7=5y = 2(1) - 7 = -5

The dimensions of the squares are (4,1)(4, 1) and (1,5)(1, -5).

Step 5

1.3.1 Make $L$ the subject of the formula: $E = \frac{1}{12}ML^2$

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Answer

To isolate LL, we first multiply both sides by 12: 12E=ML212E = ML^2

Now, divide both sides by MM: L2=12EML^2 = \frac{12E}{M}

Taking the square root: L=12EML = \sqrt{\frac{12E}{M}}

Step 6

1.3.2 Calculate the value of $L$, if $E = 8.3 \times 10^2 \, kg \, m^2$ and $M = 1.6 \times 10^3 \; kg$

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Answer

Using the formula derived: L=12(8.3×102)1.6×103L = \sqrt{\frac{12(8.3 \times 10^2)}{1.6 \times 10^3}}

Calculating: L=99.6×1021.6×103=99.61.6×10162.25×1010.79mL = \sqrt{\frac{99.6 \times 10^2}{1.6 \times 10^3}} = \sqrt{\frac{99.6}{1.6} \times 10^{-1}} \approx \sqrt{62.25 \times 10^{-1}} \approx 0.79 \, m

Step 7

1.4 Express 36 as a binary number.

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Answer

To convert 36 to binary, we can repeatedly divide by 2 and record the remainders:

  • 36 ÷ 2 = 18, remainder 0
  • 18 ÷ 2 = 9, remainder 0
  • 9 ÷ 2 = 4, remainder 1
  • 4 ÷ 2 = 2, remainder 0
  • 2 ÷ 2 = 1, remainder 0
  • 1 ÷ 2 = 0, remainder 1

Reading from bottom to top, we get 36 in binary as 1001002100100_2.

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